1948.txt

北大ACM题目例程 详细的解答过程 程序实现 算法分析

#include <stdio.h>
#include <memory.h>
#include <math.h>
#include <list>
using namespace std;
bool sign[800][800];
int l[40], n;
typedef pair<int,int> node;
node q[800*800];
int qn;
int main()
{
	int i, j, x, y, s, m;
	list< node >::iterator iter;

	scanf( "%d", &n );
	
	for( s=0,i=0; i<n; i++ )
	{
		scanf( "%d", l+i );
		s += l[i];
	}
	memset( sign, 0, sizeof( sign ) );
	sign[0][0] = 1;	
	qn = 1;
	q[0].first = 0;
	q[0].second = 0;
	for( i=0; i<n; i++ )
	{
		for( j = 0,m = qn; j < m; j++ )
		{
			x = q[j].first;
			y = q[j].second;
			if( x + l[i] <= s/2  && !sign[ x+l[i] ][y] )
			{
				sign[x+l[i]][y] = sign[y][x+l[i]] = true;
				q[qn].first = x+l[i];
				q[qn].second = y;
				qn++;
			}
			if( y + l[i] <= s/2 && !sign[ y+l[i] ][x] )
			{
				sign[y+l[i]][x] = sign[x][y+l[i]] = true;
				q[qn].first = y+l[i];
				q[qn].second = x;
				qn++;
			}
		}
	}
	double ans = -1, p, temp;
	for( i=0; i <= s/2; i++ )
	for( j=i; j <= s/2; j++ )
	if( sign[i][j] && j+i > s-(j+i) && j-i < s-(j+i) && s-j > j )
	{

		p = (double)s/2;
		if( ( temp=sqrt( p*(p-j)*(p-i)*(p-(s-i-j)) ) ) > ans )
			ans = temp;
	}
	if( ans == -1 )printf( "-1\n" );
	else printf( "%d\n", (int)(ans*100) );	
	return 0;
}